[Tom Bishop]

The difference between 45^o N and 45^o S would be very slight under the geometry of the Earth-Moon System.

[Tom Bishop]

If you were to the rear of all the observers, looking past the Earth toward the Moon, their views would vary, because of the difference in their LATITUDE. The observer at the equator will have the crescent toward the bottom of the Moon, and the observers at 45N or 45S will have it inclined to this aspect by 45 degrees each.

I don't see how that makes much sense. That doesn't use Round Earth Geometry. In fact, it seems to be assuming that the moon is very close to the earth.

[Tumeni]

If you were to the rear of all the observers, looking past the Earth toward the Moon, their views would vary, because of the difference in their LATITUDE. The observer at the equator will have the crescent toward the bottom of the Moon, and the observers at 45N or 45S will have it inclined to this aspect by 45 degrees each.

IMG

I don't see how that makes any sense at all. That doesn't use Round Earth Geometry. In fact, it is assuming that the moon is very close to the earth.

No, it is assuming you are looking PAST the Earth toward the Moon which is in the far distance. Read the legends on the diagram.

Do you look at this and think the Moon is close to my baseball?

https://imgur.com/a/Ci10Oo7

[Tumeni]

The difference between 45^o N and 45^o S would be very slight under the geometry of the Earth-Moon System.

IMG

Yes, and this corresponds to the observer on the equator looking broadly due West, the observer at 45S looking slightly North of West, and the observer at 45N looking slightly South of West.

Exactly as mooncalc predicts.

However, that's when viewed from the side of the Earth/Moon system, describing a triangle where one side is the distance between the highest and lowest observer, the other two sides their sightlines to the Moon, and the difference being a small angle at the point of the triangle at the Moon. 

If you view the situation from the rear of the observers, with you and they looking toward the Moon, the one at the equator, if you are aligned N-S, will appear to be horizontal, with the others inclined at 45 degrees each to him/her. The observer at the equator will see the crescent at the bottom of his/her Moon, the others will see the crescent inclined at 45 degrees to this - exactly as you drew in with the dotted red lines on the mooncalc screen prints, and exactly as mooncalc predicts.

[Tumeni]

The difference between 45^o N and 45^o S would be very slight under the geometry of the Earth-Moon System.

This is just my side view flipped horizontally, isn't it? You've shown Earth left, Moon right, I showed Moon left, Earth right.

Surely the easy way to do this is to calculate a triangle with two sides of 240k miles, and one of the distance between the two observers at 45N and 45S, and solve for the angle opposite the side which connects them? That would take away your totally artificial circle of the Earth going around the Moon.

The short side will be the length of chord BX, where the angle at M is 90 degrees, assuming two observers at 45N and 45S.



https://en.wikipedia.org/wiki/Chord_(geometry)

(I genuinely thought you were drawing the orbit of the Moon around the Earth. Way to make things more confusing.)

[Tom Bishop]

The difference between 45^o N and 45^o S would be very slight under the geometry of the Earth-Moon System.

https://i.imgur.com/CTOCE1I.png

This is just my side view flipped horizontally, isn't it? You've shown Earth left, Moon right, I showed Moon left, Earth right.

Surely the easy way to do this is to calculate a triangle with two sides of 240k miles, and one of the distance between the two observers at 45N and 45S, and solve for the angle opposite the side which connects them? That would take away your totally artificial circle of the Earth going around the Moon.

The short side will be the length of chord BX, where the angle at M is 90 degrees, assuming two observers at 45N and 45S.



https://en.wikipedia.org/wiki/Chord_(geometry)

I don't think it will matter. the North Pole and South Pole are still aligned with those points on the circumference with either method. I will try it out, however.

I genuinely thought you were drawing the orbit of the Moon around the Earth. Way to make things more confusing

That's okay. No hard feelings.

[model 29]

Thing is Tom, on your "close moon - flat Earth" diagram, completely different faces of the moon would be seen.  This is not observed.

Between the different latitudes on a globe however, the face of the moon rotates due to the different angles between different latitudes.  This is what is observed.

[Tom Bishop]

Thing is Tom, on your "close moon - flat Earth" diagram, completely different faces of the moon would be seen.  This is not observed.

That has been an on and off discussion. See: https://forum.tfes.org/index.php?topic=10013.0

Also, as shown on Page 1 of this thread, there are not many good sources of pictures of the moon taken simultaneously, to say what exactly happens.

Between the different latitudes on a globe however, the face of the moon rotates due to the different angles between different latitudes.  This is what is observed.

Demonstrate this with your model, rather than just saying it. I gave a demonstration. Your Round Earth model does not explain it at all.

[model 29]

there are not many good sources of pictures of the moon taken simultaneously, to say what exactly happens.

Doesn't need to be simultaneously.  If the moon were close to a flat plain, different faces would be seen as it moved overhead.  Has this ever been observed?

Your Round Earth model does not explain it at all.

It does actually.  Anyone with a ball, camera, and an object with surface features can try it.  At 45N, when viewing the moon set, there is a 90 degree difference in viewing angle from viewing it when it rises.

If you have ever seen a desktop globe in real life, or perhaps any ball, have a think about it.

[Tom Bishop]

It does actually.  Anyone with a ball, camera, and an object with surface features can try it.  At 45N, when viewing the moon set, there is a 90 degree difference in viewing angle from viewing it when it rises.

If you have ever seen a desktop globe in real life, or perhaps any ball, have a think about it.

Model it. Use the Round Earth Distances to explain it.

[Tom Bishop]

While researching this, I found several other inexplicable Round Earth items.

- The Moon is above the horizon for observers on the Prime Meridian and the Antimeridian simultaneously, when it should not be

- The Moon is pointing in the wrong direction on the horizons of the PM and AM. While the Moon will never "tilt" or "rotate" in the Round Earth model, it is possible to look at it upside-down when standing on opposite sides of the earth.

- The Moon is above the Western Horizon for observers on the Prime Meridian and the Antimeridian, when it should be on the Western Horizon for the Prime Meridian and the Eastern Horizon for the Antimeridian

Explanatory image:





Here is what the Mooncalc shows for the Prime Meridian and the Antimeridian:

Prime Meridian



Link: https://www.mooncalc.org/#/0,0,3/2018.08.12/17:58/1/2

Antimeridian



Link: https://www.mooncalc.org/#/0,180,3/2018.08.12/17:58/1/2

[Tumeni]

Tom, would you say the brown-roofed houses in this photo are close to the sign at the left of the tracks? You can see how, if you look at the Earth and Moon from behind the observers, that their lines of sight would look like the two rails, can't you?



Again, would you say the Moon here is 'close' to my baseball?





You can't just show everything as a side view and hope that will fly. If you want to show, in diagrammatic form, what something in 3D looks like, you have to show it in plan, side, front, and possibly rear views.

Diagrams to follow

[Tumeni]

If you look at the lines of sight from three observers, 45N, Equator, and 45S, then they, in plan, side-on, and underside views will look like this.



Tom, do you agree? Y/N?

If not, why not? This is the same as your diagram which indicates your difference in angle between the observers, isn't it?

[Tumeni]

This is the first step of modelling this with RE distances.



The three observers are at the points indicated by the green squares. Their lines of sight to the Moon (orange) are not to scale, the Moon would be far over to the left.

If the diameter of Earth is taken as 7926 miles, then the radius is (7926/2 =) 3959, and the length of chord C (the orange vertical) is 5598 miles, using the formula on this page ("So the length of the chord is: ")

https://www.ck12.org/trigonometry/length-of-a-chord/lesson/Length-of-a-Chord-TRIG/



If the length of this chord is 5598, and we take the distance from the 45N and 45S observers to be exactly 240k miles to the Moon, that gives us an isosceles triangle, or two right-angle triangles back-to-back, one with hypotenuse to 45N, one to 45S.
Let's solve for the angle at the Moon for the right-angles, and double it.

Right-triangle calculator;

http://www.cleavebooks.co.uk/scol/calrtri.htm

Side a = half the length of the chord, so 5598/2 = 2799, and side c = 240,000. Angle A = 0.668 degrees, so the angle at the Moon made by taking the two lines of sight from 45N and 45S is 2 * 0.668 = 1.336 degrees.

This gives us the angle at the Moon made by taking the difference between lines of sight from 45N and 45S (indicated by two orange squares).

The chord length will not vary, as long as the observers are at 45N and 45S, they are on the side of the Earth facing the Moon, and we take the distance to the Moon as a fixed 240k miles (I'm excluding any allowance for Earth's axial tilt at present, and discounting any change in the 240k as they move nearer or further to the Moon along the line of latitude)



Tom, look at the above and tell me/us - do you agree?  Y/N

Do you also agree that this is another way to calculate the angle that you were trying to calculate with your huge circle around the Moon, isn't it?

[Tumeni]

If we could look at these lines of sight from a position out in space behind the observers, not in side, plan or underside views, they will look like this




We're roughly aligned with the central observer (who we've taken as being on the equator), and the lines of sight from 45N
and 45S would start out 2799 miles away from this observer (at each end of the chord which has just been calculated), and
converge on the Moon, 240k miles away.

The Moon is decidedly not close to the baseball.

Tom, do you agree?  Y/N

 

[Tumeni]

While researching this, I found several other inexplicable Round Earth items.

- The Moon is above the horizon for observers on the Prime Meridian and the Antimeridian simultaneously, when it should not be

- The Moon is pointing in the wrong direction on the horizons of the PM and AM. While the Moon will never "tilt" or "rotate" in the Round Earth model, it is possible to look at it upside-down when standing on opposite sides of the earth.

- The Moon is above the Western Horizon for observers on the Prime Meridian and the Antimeridian, when it should be on the Western Horizon for the Prime Meridian and the Eastern Horizon for the Antimeridian

Here is what the Mooncalc shows for the Prime Meridian and the Antimeridian:

IMGs

You've shown 17.58 for both, but the Antemeridian is 12 hours ahead or behind this. This is local time.

If it's 17.58 at Greenwich, and the antemeridian point is 12 hours ahead, then it's not 17.58 there. It's 05:58 on the next day.

You've shown how the Moon looks for Greenwich, and how it looked for Australia 12 hours earlier.

Put in the correct times, and the arrows you've put on the Moon points will point from opposite directions.

You need to account for the time differences, as I did in Reply #22

[QED]

While researching this, I found several other inexplicable Round Earth items.

- The Moon is above the horizon for observers on the Prime Meridian and the Antimeridian simultaneously, when it should not be

- The Moon is pointing in the wrong direction on the horizons of the PM and AM. While the Moon will never "tilt" or "rotate" in the Round Earth model, it is possible to look at it upside-down when standing on opposite sides of the earth.

- The Moon is above the Western Horizon for observers on the Prime Meridian and the Antimeridian, when it should be on the Western Horizon for the Prime Meridian and the Eastern Horizon for the Antimeridian

Explanatory image:





Here is what the Mooncalc shows for the Prime Meridian and the Antimeridian:

Prime Meridian



Link: https://www.mooncalc.org/#/0,0,3/2018.08.12/17:58/1/2

Antimeridian



Link: https://www.mooncalc.org/#/0,180,3/2018.08.12/17:58/1/2

Thomas, you have the same time entered for both locations. You do recognize that this is actually comparing them at different times, yes?

Wow...I cannot believe you made such a flimsy error.

This is what you call research? High school students do better "research" than this.

What's wrong with you?

[Tumeni]

Do you know what you have not taken into account in your cartoon?

The Sun.

Create a proper scale-model figure which includes the thing that is illuminating the Moon in the first place (that would be the Sun, Thomas), and re-submit.

In addition, whilst he may have indicated the POSITION of the observers, he hasn't made any allowance for their orientation ...

[Tom Bishop]

Thomas, you have the same time entered for both locations. You do recognize that this is actually comparing them at different times, yes?

I didn't change the time, I just changed the location. If Mooncalc changed the time, then I wouldn't be surprised. It is an incredibly buggy web application.

Do you know what you have not taken into account in your cartoon?

The Sun.

Create a proper scale-model figure which includes the thing that is illuminating the Moon in the first place (that would be the Sun, Thomas), and re-submit.

In addition, whilst he may have indicated the POSITION of the observers, he hasn't made any allowance for their orientation ...

I've shown that it cannot be the moon that is rotating. The only out is that it is the earth that is shifting.

It doesn't seem that the Mooncalc can reproduce the Moon Tilt Illusion. I will take a look at Stellarium and compare the two.

[Curious Squirrel]

Thomas, you have the same time entered for both locations. You do recognize that this is actually comparing them at different times, yes?

I didn't change the time, I just changed the location. If Mooncalc changed the time, then I wouldn't be surprised. It is an incredibly buggy program.

If you change location you need to change the time. Why would it do that for you? That's not a buggy program, that's GIGO. 16:50 in CST is not the same time as 16:50 in GST. You need to change the time manually when you move timezones.